Answer:
Explanation:
Hello,
To find the percentage composition of muscovite mica, we'll have to first find the molecular mass of the compound.
Chemical formula = (KF)₂(Al₂O₃)₃(SiO₂)₆(H₂O)
(KF)₂ = 58.097 × 2 = 116.194g/mol
(Al₂O₃)₃ = 3 × 101.96 = 305.88g/mol
(SiO₂)₆ = 6 × 60.08 = 360.48g/mol
H₂O = 18g/mol
(KF)₂(Al₂O₃)₃(SiO₂)₆(H₂O) = 116.194 + 305.88 + 360.48 + 18 = 800.554g/mol
Potassium = (78.18 / 800.554) × 100 = 9.765%
Fluorine = (38 / 800.554) × 100 = 4.75%
Aluminium = (162 / 800.554) × 100 = 20.23%
Silicon = (168.48/800.554) × 100 = 21.04%
Oxygen = (352/800.554) × 100 = 43.97%
Hydrogen = (2 / 800.554) × 100 = 0.24%
Muscovite mica is an aluminosilicate compound or a polysillicate compound found in rocks
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?
Answer:
[tex]T2=276K[/tex]
Explanation:
Given:
Initial volume of the balloon V1 = 348 mL
Initial temperature of the balloon T1 = 255C
Final volume of the balloon V2 = 322 mL
Final temperature of the balloon T2 =
To calculate T1 in kelvin
T1= 25+273=298K
Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula
[tex](V1/T1)=(V2/T2)[/tex]
T2=( V2*T1)/V1
T2=(322*298)/348
[tex]T2=276K[/tex]
Hence, the temperature of the freezer is 276 K
Answer: 276 kelvins
Explanation:
An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as substrate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.
Answer:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.
Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin. Palmitate oxidation however, does not involve carboxylation.
Explanation:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.
Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme. The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.
Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.
The cytochromes are heme‑containing proteins that function as electron carriers in the mitochondria. Calculate the difference in the reduction potential (ΔE∘′) and the change in the standard free energy (ΔG∘′) when the electron flow is from the carrier with the lower reduction potential to the higher. cytochrome c1 (Fe3+)+e−↽−−⇀cytochrome c1 (Fe2+)E∘′=0.22 V cytochrome c (Fe3+)+e−↽−−⇀cytochrome c (Fe2+)E∘′=0.254 V Calculate ΔE∘′ and ΔG∘′ .
Complete Question
The complete question is shown on the first uploaded image
Answer:
The change in reduction potential is [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]
The change in standard free energy is [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]
Explanation:
From the question we are told that
At the anode
[tex]cytochrome \ c_1 \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c_1 \ (Fe^{2+}) \ \ E^o = 0.22 \ V[/tex]
At the cathode
[tex]cytochrome \ c \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c \ (Fe^{2+}) \ \ E^o = 0.254 \ V[/tex]
The difference in the reduction potential is mathematically represented as
[tex]\Delta E^o = E^o_{cathode} - E^o_{anode}[/tex]
substituting values
[tex]\Delta E^o = 0.254 - 0.220[/tex]
[tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]
The change in the standard free energy is mathematically represented as
[tex]\Delta G^o = -n * F * E^o_{cell}[/tex]
Where F is the Faraday constant with value F = 96485 C
and n i the number of the number of electron = 1
So
[tex]\Delta G^o = -(1) * 96485 * 0.034[/tex]
[tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]
The iceman known as Otzi was discovered on a mountain on the Austrian-Italian border. Samples of his hair and bones had carbon-14 activity that was about 12.5% of that present in new hair or bone. How long ago did Otzi live if the half-life for C-14 is 5730 years
Answer:
1432.5 years
Explanation:
The rate of decay of a radioactive isotope is the characteristics of the isotope and it is usually expressed in terms of its half-life.
The half-life of a radioactive element is the time taken for half of the total number of atoms in a given sample of the element to decay or the time taken for the intensity of radiation to fall to half of its original value.
From the given question.
Since the same of his bones had a carbon-14 activity that was about 12.5% of that present in new hair or bone.
Thus; the time taken to reduce the amount of the sample to one-quarter of its amount(12.5%) = the half life for C-14 (5730 years)
The time taken for how long Otzi live = 5730/4 = 1432.5 years
Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.800 pint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?
Answer:
[tex]\large \boxed{\text{122 000 J}}[/tex]
Explanation:
1. Calculate the energy needed
[tex]\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}} = \text{ 29 120 cal}[/tex]
2. Convert calories to joules
[tex]\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}[/tex]
Which of the following aqueous solutions are good buffer systems?
0.31 M ammonium bromide + 0.39 M ammonia
0.31 M nitrous acid + 0.25 M potassium nitrite
0.21 M perchloric acid + 0.21 M potassium perchlorate
0.16 M potassium cyanide + 0.21 M hydrocyanic acid
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite
0.13 M nitrous acid + 0.12 M potassium nitrite
0.15 M potassium hydroxide + 0.22 M potassium bromide
0.23 M hydrobromic acid + 0.20 M potassium bromide
0.34 M calcium iodide + 0.29 M potassium iodide
0.33 M ammonia + 0.30 M sodium hydroxide
0.20 M nitrous acid + 0.18 M potassium nitrite
0.30 M ammonia + 0.34 M ammonium bromide
0.29 M hydrobromic acid + 0.22 M sodium bromide
0.17 M calcium hydroxide + 0.28 M calcium bromide
0.34 M potassium iodide + 0.27 M potassium bromide
Answer:
Answers are in the explanation.
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:
0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.
0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.
0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.
0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.
0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.
0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.
0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.
0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.
0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.
0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.
0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.
0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.
0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.
0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.
The bromine test (part d) is often used as an indication of unsaturation(double and triple bonds). Explain why your result for trichloroethylene and toluene were different than for the simple alkene produc
Answer:
Toluene is an aromatic compound not an alkene
Bromine test is used to determine the presence of unsaturation in the given compound. The trichloroethylene does not have any unsaturation while toluene have double bonds of benzene ring. Therefore, the Bromine test can differentiate between trichloroethylene and toluene.
What is degree of unsaturation?The degree of unsaturation of an organic compounds can be categorised two types: saturated and unsaturated. Saturated compounds are those that have only single bonds. An unsaturated compound are those that has a double bond, triple bond, and/or ring(s).
The alkanes with only single bonds are classified as saturated whereas the alkenes and alkynes with double and triple bonds are classified as unsaturated hydrocarbons.
The degree of unsaturation formula helps in finding whether a compound is saturated or unsaturated.
In the Bromine test when the bromine solution will be added into the compound if the brown color of the solution will disappear it means the unsaturation is present in the given compound.
Therefore, the we can distinguish between trichloroethylene and toluene with bromine test.
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Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:
Here is the complete question.
Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of: the following properties. (Use the notation >, <, or =, for example B=C>A.)
(a) pressure
(b) average molecular kinetic energy
(c) diffusion rate after the valve is opened
(d) total kinetic energy of the molecules
Answer:
Explanation:
Given that:
Three flask A,B, C:
contains a volume of 8-L
mass m = 4g &;
Temperature = 276 K
Flask A = He
Flask B = H₂
Flask C = CH₄
a) From the ideal gas equation:
PV = nRT
where;
n = number of moles = mass (m)/molar mass (mm)
Then:
PV = m/mm RT
If T ,m and V are constant for the three flasks ; then
P ∝ 1/mm
As such ; the smaller the molar mass the larger the pressure.
Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂, He has an higher molecular weight .
Then the order of pressure in the flask is :
[tex]\mathbf{P_B >P_A>P_C}[/tex]
where :
[tex]P_A[/tex] = pressure in the flask A
[tex]P_B[/tex] = pressure in the flask B
[tex]P_C[/tex]= Pressure in the flask C
b)
average molecular kinetic energy
We all know that the average molecular kinetic energy varies directly proportional to the temperature.
Thus; the given temperature = 276 K
∴ The order of the average molecular kinetic energy is [tex]\mathbf{K.E_A =K.E_B =K.E_C}[/tex]
c)
The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.
So;
rate of diffusion ∝ [tex]\dfrac{1}{\sqrt{mm} }[/tex]
where;
[tex]D_A[/tex] = rate of diffusion in flask A
[tex]D_B[/tex] = rate of diffusion in flask B
[tex]D_C[/tex] = rate of diffusion in flask C
Thus; the order of the rate of diffusion = [tex]D_B[/tex] > [tex]D_A[/tex] > [tex]D_C[/tex]
d) total kinetic energy of the molecules .
The kinetic energy deals with how the speed of particles of a substance determines how fast the substances will diffuse in a given set of condition.
The the order of the total kinetic energy depends on the molecular speed
Thus; the order of the total kinetic energy for the three flask is as follows:
[tex]\mathbf{ K.E_B>K.E_A>K.E_C}[/tex]
In supersonic flights, molecules break apart and react chemically. which safety features protect the plane?
Answer:
Explanation:
The heat Shield are materials (usually made of metals) protect us from heat by absoring lots of heat and gradually releasing heat by surrounding air cirucaltion
Answer:
Heat shield
Explanation: Most heat shields consist of one or more layers of stamped metal that are shaped into a shield that is designed to wrap around the exhaust manifold. The shield acts as a barrier and heat sink, preventing the heat from the manifold from reaching any of the components under the hood and potentially causing damage.
An electron in a 3s3s orbital penetrates into the region occupied by core electrons more than electrons in a 3p3p orbital. An electron in a orbital penetrates into the region occupied by core electrons more than electrons in a orbital. true false
Answer:
True
Explanation:
The penetrating ability of electrons in the orbitals is in the order s > p > d > f
An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.
Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for water.) Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/ mL for water.)
Answer: The freezing point and boiling point of the solution are [tex]-6.6^0C[/tex] and [tex]101.8^0C[/tex] respectively.
Explanation:
Depression in freezing point:
[tex]T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_f[/tex] = freezing point of solution = ?
[tex]T^o_f[/tex] = freezing point of water = [tex]0^0C[/tex]
[tex]k_f[/tex] = freezing point constant of water = [tex]1.86^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_f=-6.6^0C[/tex]
Therefore,the freezing point of the solution is [tex]-6.6^0C[/tex]
Elevation in boiling point :
[tex]T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of water = [tex]100^0C[/tex]
[tex]k_b[/tex] = boiling point constant of water = [tex]0.52^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_b=101.8^0C[/tex]
Thus the boiling point of the solution is [tex]101.8^0C[/tex]
which proess is part of the carbon cycle
Answer:
The key processes in the carbon cycle are: carbon dioxide from the atmosphere is converted into plant material in the biosphere by photosynthesis.
Explanation:
organisms in the biosphere obtain energy by respiration and so release carbon dioxide that was originally trapped by photosynthesis. ... The carbon becomes part of the .
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.9 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.
Answer:
The correct answer to the following question will be "90.6 kJ/mol".
Explanation:
The total reactant solution will be:
[tex](31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g[/tex]
The produced energy will be:
[tex]=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K[/tex]
[tex]=450.35\times 3.2[/tex]
[tex]=1441.12 \ J[/tex]
The reaction will be:
⇒ [tex]HCl+NaOH \rightarrow NaCl+H_{2}O[/tex]
Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.
Therefore, the moles generated by NaCl will indeed be:
= [tex](\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}[/tex]
= [tex]0.0318\times 0.500[/tex]
= [tex]0.0159 \ mole \ of \ NaCl[/tex]
Now,
= [tex]\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}[/tex]
= [tex]906364.7[/tex]
= [tex]90.6 \ KJ/mol \ NaCl[/tex]
How many moles of each product form when the given amount of each reactant completely reacts. C3H8(g)+5O2yields 3CO2(g)+4H2O(g). 4.6 moles of C3H8
Answer: 13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced
Explanation:
The given balanced reaction is;
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
Given : 4.6 moles of [tex]C_3H_8[/tex]
According to stoichiometry :
1 mole of [tex]C_3H_8[/tex] give = 3 moles of [tex]CO_2[/tex]
Thus 4.6 moles of [tex]C_3H_8[/tex] will give =[tex]\frac{3}{1}\times 4.6=13.8moles[/tex] of [tex]CO_2[/tex]
1 mole of [tex]C_3H_8[/tex] give = 4 moles of [tex]H_2O[/tex]
Thus 4.6 moles of [tex]C_3H_8[/tex] give =[tex]\frac{4}{1}\times 4.6=18.4moles[/tex] of [tex]H_2O[/tex]
Thus 13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced from the given moles of reactant [tex]C_3H_8[/tex]
The reaction of 15 moles carbon with 30 moles O2 will
result in a theoretical yield of __ moles CO2.
Answer:
15 moles.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]C+O_2\rightarrow CO_2[/tex]
Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:
[tex]n_{CO_2}=15molC*\frac{1molCO_2}{1molC} \\\\n_{CO_2}=15molCO_2[/tex]
Best regards.
Look at the picture and observations below.
Observations: The bee's wings are moving very fast.
The bee's wings are much smaller than its body.
what’s the answer ?
Answer:
How are bees able to fly?
Explanation:
Which describes an effect that ocean currents have on short-term climate change? Ocean currents increase the strength of prevailing winds, which can cool the air and land. Ocean currents can carry cold water, which can cool the air and land. Ocean currents increase hurricane activity, which can raise the temperature of the air and land. Ocean currents can carry warm water, which causes hurricane activity and raises the temperature of the air and land.
Answer: B
Ocean currents can carry cold water, which can cool the air and land.
Explanation:
eet ees wat eet ees
plz mark brainliest
Answer:
B is right
Explanation:
Calculate the temperature and state the appropriate phase(s) (solid, liquid, vapor) for each substance: This temperature is: Fahrenheit A. SubstanceMelting Point (K)Boiling Point (K)Phase(s) Oxygen, O254.7590.19 Methane, CH493.15109.10 Water, H2O273.15373.15
Answer:
Explanation:
At 54.75K melting point, Oxygen is in gas (vapour) phase
At 373.15K boiling point, water is in liquid phase.
At 109.10K boiling point methane is in gas (vapour) phase.
describe how would you use chromatography to show whether blue ink contains a single purple dye or a mixture of dyes
Explanation:
if the solution placed on the chromatography is pure there will be formation of one spot from the baseline and will go farthest to the front line unlike the impure one
With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water, and the different colors of the ink are revealed as the water moves up the paper. All of this is made possible by the water base and variety of salabilities or densities that make up ink.
What is chromatography ?Separating mixture's constituent parts by chromatography is a method. The mixture is dissolved in a material known as the mobile phase to start the process, which then transports it through a material known as the stationary phase.
A little dot of the ink to be separated is placed at one end of a strip of filter paper to perform ink chromatography. The paper strip's opposite end is submerged in a solvent. The solvent moves up the paper strip, dissolving the chemical combination as it goes and pulling it up the paper.
Throughout the experiment, the dyes are pulled along by the mobile phase (water) as it gently advances up the stationary phase (paper).
Thus, With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water.
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Draw structural formulas for all the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an E2 mechanism.
Answer:
Explanation:
Kindly note that I have attached the complete question as an attachment.
Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.
The possible products are as follows;
Please check attachment for complete equations and diagrams of compounds too.
s the following nuclear equation balanced? yes no
Answer:
Yes.
Explanation:
The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.
Therefore, the equation will be reduced to:
226 - 4 = 222
88 - 2 = 86
Hence, the equation is balanced.
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compressor. Determine the compressor power per unit mass flow rate if the device is (a) isentropic, (b) polytropic with n =1.3, (c) isothermal
Answer:
(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:
[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]
[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]
Next, we compute the final temperature:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]
Thus, the work is computed by:
[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]
And the isentropic work:
[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) Finally, for isothermal, final temperature is not required as it could be computed as:
[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Regards.
Why need to add NaAlF6 to Al2O3?
How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
O A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases
C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.
Answer:
A
Explanation:
State the effect of anion hydrolysis on the pH of water
Answer:
Depending on the anions and cations present within a hydrolysis reaction, the solution can be more... ... This lesson will explain how this occurs. ... that could react with water and create products that affect the characteristics of the solution.
Answer:
Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.
Explanation:
I hope this is the answer your looking for
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M
Answer:
31.652g of Na3PO4
Explanation:
We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:
Na3PO4 will dessicate in solution as follow:
Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)
From the balanced equation above,
1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.
Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e
xM Na3PO4 = (1.10 x 1)/3
xM Na3PO4 = 0.367M
Therefore, the molarity of Na3PO4 is 0.367M.
Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:
Molarity of Na3PO4 = 0.367M
Volume = 525mL = 525/1000 = 0.525L
Mole of Na3PO4 =..?
Molarity = mole /Volume
0.367 = mole /0.525
Cross multiply
Mole of Na3PO4 = 0.367 x 0.525
Mole of Na3PO4 = 0.193 mole.
Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:
Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol
Mole of Na3PO4 = 0.193 mole
Mass of Na3PO4 =.?
Mass = mole x molar mass
Mass of Na3PO4 = 0.193 x 164
Mass of Na3PO4 = 31.652g
Therefore, 31.652g of Na3PO4 is needed to prepare the solution.
Consider the insoluble compound zinc carbonate , ZnCO3 . The zinc ion also forms a complex with hydroxide ions . Write a balanced net ionic equation to show why the solubility of ZnCO3 (s) increases in the presence of hydroxide ions and calculate the equilibrium constant for this reaction. For Zn(OH)42- , Kf = 2.9×1015 . Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
The net ionic equation is [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]
The equilibrium constant is [tex]K = 4.06 *10^{4}[/tex]
Explanation:
From the question we are that
The [tex]K_f = 2.9 *10^{15 }[/tex]
The ionic equation is chemical represented as
Step 1
[tex]ZnCO_3 _{(s)}[/tex] ⇔ [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex] The solubility product constant for stage is [tex]K_{sp} = 1.4*10^{-11}[/tex]
Step 2
[tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex] ⇔ [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex] The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]
The net reaction is
[tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]
The equilibrium constant is mathematically evaluated as
[tex]K = K_{sp} * K_f[/tex]
substituting values
[tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]
[tex]K = 4.06 *10^{4}[/tex]
When ethanol, C2H5OH (a component in some gasoline mixtures) is burned in air, one molecule of ethanol combines with three oxygen molecules to form two CO2 molecules and three H2O molecules.
A) Write the balanced chemical equation for the reaction described.
B) How many molecules of CO2 and H2O would be produced when 2 molecules ethanol are consumed? Equation?
C) How many H2O molecules are formed, then 9 O2 molecules are consumed? What conversion factor did you use? Explain!
D) If 15 ethanol molecules react, how many molecules O2 must also react? What conversion factor did you use? Explain!
Answer:
1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)
2) four molecules of CO2 will be produced and six molecules of water
3)9 molecules of water are formed when 9 molecules of oxygen are consumed.
4) 45 molecules of oxygen
Explanation:
The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;
C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)
Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;
2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)
Hence four molecules of CO2 are formed and six molecules of water are formed
From the balanced stoichiometric equation;
3 molecules of oxygen yields 3 molecules of water
Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water
Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.
From the reaction equation;
1 molecule of ethanol reacts with 3 molecules of oxygen
Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen
How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases.
O C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.
A scientist measures the standard enthalpy change for the following reaction to be -115.5 kJ: CO(g) + Cl2(g)___COCl2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of COCl2(g) is ________ kJ/mol.
Answer:
-226.0kJ = ΔH°f COCl₂(g)
Explanation:
Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.
For the reaction:
CO(g) + Cl₂(g) → COCl₂(g)
Hess's law is:
ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))
ΔH°f CO(g) is -110.5kJ/mol
ΔH°f Cl₂(g) is 0 kJ/mol
Replacing in Hess's law:
-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)
-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ
-226.0kJ = ΔH°f COCl₂(g)